重邮 信号与系统考试答案 考研必备

r2 2r 2 0 r1,2 1 jyx(k) Cx1( 1 j)k Cx2( 1 j)k yx(0) Cx1 Cx2 0

yx(1) ( 1 j)Cx1 ( 1 j)Cx2 1jj

Cx1 ,Cx2

22

jj

yx(k) ( )( 1 j)k ( 1 j)k

223 jk3 k

(2)e2 (2)ksink

4

k 0

(3)y(k+2)+2y(k+1)+y(k)=0 yx(0) yx(1) 1 解：

r2 2r 1 0 (r 1)2 0 r1 r2 1

yx(k) (Cx1 Cx2K)( 1)k

yx(0) Cx1 1

yx(1) (Cx1 Cx2)( 1) 1

Cx1 1

C 2 x2

yx(k) (1 2k)( 1)k k 0

(4)

y(k) 2y(k 1) 0 yx(0) 2

k解： 2 0 2 yx(k) Cx(2)

yx(0) Cx 2 故 yx(k) 2(2)k k>=0

(5)

y(k) 2y(k 1) 4y(k 2) 0 yx(0) 0,yx(1) 2

解：

2 2 4 0 即 ( 1)2 3 0

特征根 1,2 1 3j

yx(k) Cx1( 1 j)k Cx2( 1 j)k